j_shreyans wrote:A driver paid n dollars for auto insurance for the year 1997. This annual premium was raised by p percent for the year 1998; for each of the years 1999 and 2000, the premium was decreased by 1/6 from the previous year's figure. If the driver's insurance premium for the year 2000 was again n dollars, what is the value of p?
A)12
B)100/3
C)36
D)44
E)50
Here's an algebraic solution.
First recognize that increasing a value by p percent is the same as multiplying by
(100 + p)/100
For example, increasing a value by 7% is the same as multiplying by (100 + 7)/100
[i.e. 1.07]
Likewise, increasing a value by 25% is the same as multiplying by (100 + 25)/100
[i.e. 1.25]
Also, recognize that decreasing a value by 1/6 is the same as multiplying by
5/6
Okay, let's begin.
Premium for 1997: n
Premium for 1998: [(100 + p)/100]n
Premium for 1999: (5/6)[(100 + p)/100]n
Premium for 2000: (5/6)(5/6)[(100 + p)/100]n
We're told that the premium for 2000 is n.
So, we can write: (5/6)(5/6)[(100 + p)/100]n = n
Divide both sides by n to get: (5/6)(5/6)[(100 + p)/100] = 1
Simplify: (25/36)[(100 + p)/100] = 1
Simplify: (25)(100 + p)/(36)(100) = 1
Simplify: (100 + p)/(36)(4) = 1
So we know that 100 + p = (36)(4)
Simplify: 100 + p = 144
So, p =
44
Answer:
D
Cheers,
Brent